2018暑期多校1

Posted 2020-12-24 -xiangyang

1001

Problem Description

Given an integer n, Chiaki would like to find three positive integers x, y and z such that: n=x+y+z, x∣n, y∣n, z∣n and xyz is maximum.

 

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤106), indicating the number of test cases. For each test case:
The first line contains an integer n (1≤n≤106).

 Output

For each test case, output an integer denoting the maximum xyz. If there no such integers, output −1 instead.

 Sample Input

3

1

2

3

 Sample Output

-1

-1

1

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int T;
 6     scanf("%d",&T);
 7     while (T--)
 8     {
 9         long long n;
10         scanf("%lld",&n);
11         if(n%3==0)
12         {
13             printf("%lld
",(n/3)*(n/3)*(n/3));
14         } else if(n%4==0)
15         {
16             printf("%lld
",(n/2)*(n/4)*(n/4));
17         }
18         else printf("-1
");
19 
20 
21     }
22     return 0;
23 }

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HDU 6301

Problem Description

Chiaki has an array of n positive integers. You are told some facts about the array: for every two elements ai and aj in the subarray al..r (l≤i<j≤r), ai≠ajholds.
Chiaki would like to find a lexicographically minimal array which meets the facts.

 

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains two integers n and m (1≤n,m≤105) -- the length of the array and the number of facts. Each of the next m lines contains two integers li and ri (1≤li≤ri≤n).

It is guaranteed that neither the sum of all n nor the sum of all m exceeds 106.

 

 

Output

For each test case, output n integers denoting the lexicographically minimal array. Integers should be separated by a single space, and no extra spaces are allowed at the end of lines.

 

 

Sample Input

3

2 1

1 2

4 2

1 2

3 4

5 2

1 3

2 4

 

 

Sample Output

1 2

1 2 1 2

1 2 3 1 1

 

 题解：

先对给定的区间排序，然后每次讲前一个比后一个区间多出的部分加入优先对列，找到最小的然后放到ans数组。这个代码有点卡时间，

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define LL long long
 4 
 5 const int MAXN = 1e5+10;
 6 
 7 
 8 int t,n,m;
 9 
10 int ans[MAXN];
11 struct node{
12     int l,r;
13 }p[MAXN],pre;
14 struct CMP{
15     bool operator ()(int a,int b){
16         return a > b;
17     }
18 };
19 bool cmp(node a,node b){
20     if(a.l!=b.l)    return a.l < b.l;
21     else return a.r > b.r;
22 }
23 
24 int main(){
25     scanf("%d",&t);
26     while(t--){
27         memset(ans,0,sizeof(ans));
28         memset(p,0,sizeof(p));
29         scanf("%d%d",&n,&m);
30         priority_queue<int, vector<int>,CMP>Q;
31         for(int i=0;i<m;i++){
32             scanf("%d%d",&p[i].l,&p[i].r);
33         }
34         sort(p,p+m,cmp);
35         int maxx = 1;
36         pre.l = p[0].l;
37         pre.r = p[0].r;
38         for(int i=1;i<=p[0].r-p[0].l+1;i++){
39             ans[p[0].l+i-1] = i;
40             maxx = max(maxx,i);
41         }
42         for(int i=1;i<m;i++){
43             if(p[i].r <= pre.r) continue;
44             else {
45                 for(int j=pre.l;j<min(p[i].l,pre.r+1);j++){
46                     Q.push(ans[j]);
47                 }
48                 int temp =max(pre.r+1,p[i].l);
49                 while(!Q.empty()&&temp<=p[i].r){
50                     ans[temp++]=Q.top();
51                     Q.pop();
52                 }
53                 if(temp<=p[i].r){
54                     for(int k=temp;k<=p[i].r;k++){
55                         ans[k]=++maxx;
56                     }
57                 }
58             }
59             pre.l=p[i].l;
60             pre.r=p[i].r;
61         }
62         for(int i=1;i<=n;i++){
63             if(ans[i]==0)ans[i]=1;
64         }
65         for(int i=1;i<n;i++){
66             printf("%d ",ans[i]);
67         }printf("%d
",ans[n]);
68     }
69     return 0;
70 }

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